3.343 \(\int (b \cos (c+d x))^{2/3} (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^3(c+d x) \, dx\)

Optimal. Leaf size=145 \[ \frac{3 A b^2 \sin (c+d x)}{4 d (b \cos (c+d x))^{4/3}}-\frac{3 (A+4 C) \sin (c+d x) (b \cos (c+d x))^{2/3} \, _2F_1\left (\frac{1}{3},\frac{1}{2};\frac{4}{3};\cos ^2(c+d x)\right )}{8 d \sqrt{\sin ^2(c+d x)}}+\frac{3 b B \sin (c+d x) \, _2F_1\left (-\frac{1}{6},\frac{1}{2};\frac{5}{6};\cos ^2(c+d x)\right )}{d \sqrt{\sin ^2(c+d x)} \sqrt [3]{b \cos (c+d x)}} \]

[Out]

(3*A*b^2*Sin[c + d*x])/(4*d*(b*Cos[c + d*x])^(4/3)) + (3*b*B*Hypergeometric2F1[-1/6, 1/2, 5/6, Cos[c + d*x]^2]
*Sin[c + d*x])/(d*(b*Cos[c + d*x])^(1/3)*Sqrt[Sin[c + d*x]^2]) - (3*(A + 4*C)*(b*Cos[c + d*x])^(2/3)*Hypergeom
etric2F1[1/3, 1/2, 4/3, Cos[c + d*x]^2]*Sin[c + d*x])/(8*d*Sqrt[Sin[c + d*x]^2])

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Rubi [A]  time = 0.192641, antiderivative size = 145, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.098, Rules used = {16, 3021, 2748, 2643} \[ \frac{3 A b^2 \sin (c+d x)}{4 d (b \cos (c+d x))^{4/3}}-\frac{3 (A+4 C) \sin (c+d x) (b \cos (c+d x))^{2/3} \, _2F_1\left (\frac{1}{3},\frac{1}{2};\frac{4}{3};\cos ^2(c+d x)\right )}{8 d \sqrt{\sin ^2(c+d x)}}+\frac{3 b B \sin (c+d x) \, _2F_1\left (-\frac{1}{6},\frac{1}{2};\frac{5}{6};\cos ^2(c+d x)\right )}{d \sqrt{\sin ^2(c+d x)} \sqrt [3]{b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(b*Cos[c + d*x])^(2/3)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^3,x]

[Out]

(3*A*b^2*Sin[c + d*x])/(4*d*(b*Cos[c + d*x])^(4/3)) + (3*b*B*Hypergeometric2F1[-1/6, 1/2, 5/6, Cos[c + d*x]^2]
*Sin[c + d*x])/(d*(b*Cos[c + d*x])^(1/3)*Sqrt[Sin[c + d*x]^2]) - (3*(A + 4*C)*(b*Cos[c + d*x])^(2/3)*Hypergeom
etric2F1[1/3, 1/2, 4/3, Cos[c + d*x]^2]*Sin[c + d*x])/(8*d*Sqrt[Sin[c + d*x]^2])

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

Rule 3021

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m +
1)*(a^2 - b^2)), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(a*A - b*B + a*
C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e,
f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 2643

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1)*Hypergeomet
ric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2])/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]), x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rubi steps

\begin{align*} \int (b \cos (c+d x))^{2/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx &=b^3 \int \frac{A+B \cos (c+d x)+C \cos ^2(c+d x)}{(b \cos (c+d x))^{7/3}} \, dx\\ &=\frac{3 A b^2 \sin (c+d x)}{4 d (b \cos (c+d x))^{4/3}}+\frac{3}{4} \int \frac{\frac{4 b^2 B}{3}+\frac{1}{3} b^2 (A+4 C) \cos (c+d x)}{(b \cos (c+d x))^{4/3}} \, dx\\ &=\frac{3 A b^2 \sin (c+d x)}{4 d (b \cos (c+d x))^{4/3}}+\left (b^2 B\right ) \int \frac{1}{(b \cos (c+d x))^{4/3}} \, dx+\frac{1}{4} (b (A+4 C)) \int \frac{1}{\sqrt [3]{b \cos (c+d x)}} \, dx\\ &=\frac{3 A b^2 \sin (c+d x)}{4 d (b \cos (c+d x))^{4/3}}+\frac{3 b B \, _2F_1\left (-\frac{1}{6},\frac{1}{2};\frac{5}{6};\cos ^2(c+d x)\right ) \sin (c+d x)}{d \sqrt [3]{b \cos (c+d x)} \sqrt{\sin ^2(c+d x)}}-\frac{3 (A+4 C) (b \cos (c+d x))^{2/3} \, _2F_1\left (\frac{1}{3},\frac{1}{2};\frac{4}{3};\cos ^2(c+d x)\right ) \sin (c+d x)}{8 d \sqrt{\sin ^2(c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.237174, size = 123, normalized size = 0.85 \[ -\frac{3 \sqrt{\sin ^2(c+d x)} \csc (c+d x) \sec ^2(c+d x) (b \cos (c+d x))^{2/3} \left (2 \cos (c+d x) \left (C \cos (c+d x) \, _2F_1\left (\frac{1}{3},\frac{1}{2};\frac{4}{3};\cos ^2(c+d x)\right )-2 B \, _2F_1\left (-\frac{1}{6},\frac{1}{2};\frac{5}{6};\cos ^2(c+d x)\right )\right )-A \, _2F_1\left (-\frac{2}{3},\frac{1}{2};\frac{1}{3};\cos ^2(c+d x)\right )\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Cos[c + d*x])^(2/3)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^3,x]

[Out]

(-3*(b*Cos[c + d*x])^(2/3)*Csc[c + d*x]*(-(A*Hypergeometric2F1[-2/3, 1/2, 1/3, Cos[c + d*x]^2]) + 2*Cos[c + d*
x]*(-2*B*Hypergeometric2F1[-1/6, 1/2, 5/6, Cos[c + d*x]^2] + C*Cos[c + d*x]*Hypergeometric2F1[1/3, 1/2, 4/3, C
os[c + d*x]^2]))*Sec[c + d*x]^2*Sqrt[Sin[c + d*x]^2])/(4*d)

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Maple [F]  time = 0.41, size = 0, normalized size = 0. \begin{align*} \int \left ( b\cos \left ( dx+c \right ) \right ) ^{{\frac{2}{3}}} \left ( A+B\cos \left ( dx+c \right ) +C \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*cos(d*x+c))^(2/3)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3,x)

[Out]

int((b*cos(d*x+c))^(2/3)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac{2}{3}} \sec \left (d x + c\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))^(2/3)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3,x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c))^(2/3)*sec(d*x + c)^3, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac{2}{3}} \sec \left (d x + c\right )^{3}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))^(2/3)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3,x, algorithm="fricas")

[Out]

integral((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c))^(2/3)*sec(d*x + c)^3, x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))**(2/3)*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**3,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac{2}{3}} \sec \left (d x + c\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))^(2/3)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3,x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c))^(2/3)*sec(d*x + c)^3, x)